Answer
$ y= -4 x^2+\frac{50}{3} x+6 $
Work Step by Step
Since the zeros of the function are located at $x=- \frac{1}{5}$ and $x= \frac{9}{2}$, we can write
$$\begin{aligned}
y & =a\left(x+\frac{1}{3}\right)\left(x-\frac{9}{2}\right) \\
& =a\left[\left(x+\frac{1}{3}\right) x-\frac{9}{2}\left(x+\frac{1}{3}\right)\right] \\
& =a\left[x^2+\frac{1}{3} x-\frac{9}{2} x-\frac{9}{2} \cdot \frac{1}{3}\right] \\
& =a\left[x^2+\frac{2 x-9 \cdot 3 x}{6}-\frac{3}{2}\right] \\
& =a\left(x^2-\frac{25}{6} x-\frac{3}{2}\right).
\end{aligned}
$$ Given that $f(3)=20$, we can use the point $(x,y) = (3 , 20) $ to find the constant $a$.
$$
\begin{aligned}
a\left(3+\frac{1}{3}\right)\left(3-\frac{9}{2}\right) & =20 \\
a\left(\frac{3 \cdot 3+1}{3}\right)\left(\frac{3 \cdot 2-9}{2}\right) & =20 \\
a\left(\frac{10}{3}\right)\left(-\frac{3}{2}\right) & =20 \\
-5 a & =20 \\
a & =-\frac{20}{5} \\
& =-4.
\end{aligned}
$$ This gives $$
\begin{aligned}
\Rightarrow y & =-4 x^2-\frac{25}{6}(-4) x-\frac{3}{2}(-4) \\
& =-4 x^2+\frac{50}{3} x+6.
\end{aligned}
$$
